Answer
$|f(x,y) -f(0,0) | \lt \epsilon $
Work Step by Step
We have $f(x,y)=\dfrac{x+y}{x^2+1}$ and $f(0,0)=0$
Now, $|f(x,y) -f(0,0) | \lt \epsilon \implies |\dfrac{x+y}{x^2+1}-0| \lt 0.01 $
This implies that $\dfrac{|x+y|}{x^2+1} \lt 0.01$
Since,$|x+y| \leq 2 \sqrt {x^2+y^2 }; x^2+1 \geq 1$
So, $\dfrac{|x+y|}{x^2+1} \leq 2 \sqrt {x^2+y^2}$
This implies that $2 \sqrt {x^2+y^2} \lt 0.01 \implies \sqrt {x^2+y^2} \lt 0.005$
Suppose $\delta =0.005 \implies \sqrt {x^2+y^2 } \lt \delta$
Thus, $|f(x,y) -f(0,0) | \lt \epsilon $
